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murzikaleks [220]
4 years ago
10

How many significant figures does 0.0006510 of have?

Chemistry
2 answers:
arsen [322]4 years ago
7 0
It has 4 significant figures. If you see the Zero after the one decimal point you don’t count that and instead just started at 6
torisob [31]4 years ago
4 0

There are 4 sig. Fig.s in 0.0006510, the 6510 are significant and the 0.000 are insignificant,  as zeros can not lead and the are not captive zeros, and have no non-zero digit leading them; they are seen as insignificant zeros and not counted.

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Which statement best describes chemical properties of matter? Chemical properties, such as density, must be observed when a subs
alisha [4.7K]
Chemical properties of matter can only be observed and measured by performing a chemical change.
- Density and boiling are not chemical changes.
- This leaves reactivity and combustion. Reactivity does not need to be observed at STP so the statement is wrong.

Combustion is the best answer. 

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Sodiun atom, potassium atom, and cesium atoms have the same ?
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Sodium atom , potassium atom and cesium atom have the same group number which is group 1
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2CO + O2  2CO2
murzikaleks [220]

Answer:

75 L Co2

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Explanation:

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3 0
3 years ago
A 0.4657 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr b
Yanka [14]

Answer:

The mass percentage of bromine in the original compound is 81,12%

Explanation:

<u>Step 1: Calculate moles AgBr</u>

moles AgBr = mass AgBr / molar mass  AgBr

= 0.8878 g / 187.77 g/mol

= 0.00472812 moles AgBr

⇒

Since 1 mol AgBr contains 1 mol Br-

Then the amount of moles Br- in the original sample must also have  been 0.00472812 moles

<u>Step 2:</u> Calculating mass Br-

mass Br- = molar mass Br x moles  Br-

= 79.904 g/mol x 0.00472812 mol

= 0.377796 g Br-

⇒

There were 0.377796 g Br- in the original sample

<u>Step 3:</u> Calculating mass percentage Br-

⇒mass percentage  = actual mass Br- / total mass x 100%

% mass Br = 0.377796 g / 0.4657 g x 100  %

= 81.12%

7 0
3 years ago
The argon atoms are excited into an excited state before emitting the 488.0 nm laser. It is known that the energy of the first i
Jobisdone [24]

Answer:

Explanation:

Given that:

The argon atoms are excited into an excited state before emitting the 488.0 nm laser.

the energy of the first ionization energy of argon is 1520 kJ mol-1.

SInce 1 eV = 96.49 kJ/mol

Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV

=  15.75 eV

To find where  the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.

E = \dfrac{hc}{\lambda}

E  = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}

E  = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}

E  = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}

E  =4.057 \times 10^{-19} \ J

Converting Joules (J) to eV ; we get,

E  =\dfrac{4.057 \times 10^{-19}}{1.6 \times 10^{-19}}

E = 2.53 eV

The energy levels of the first exited state = -13.223 eV

8 0
4 years ago
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